Pls help me i’ll really appreciate it!!

6.184×105=

9966 [12]

Answer:

649.32

Step-by-step explanation:

7 0

4 years ago

Paul and jose are trying to measure the height of a tree. paul is standing 19m from the foot of the tree and measures the angle

Nina [5.8K]

The firts thig we are going to do is create tow triangles using the angles of elevation of Paul and Jose. Since the problem is not giving us their height we'll assume that the horizontal line of sight of both of them coincide with the base of the tree.
We know that Paul is 19m from the base of the tree and its elevation angle to the top of the tree is 59°. We also know that the elevation angle of Jose and the top of the tree is 43°, but we don't know the distance between Paul of Jose, so lets label that distance as $x$ .
Now we can build a right triangle between Paul and the tree and another one between Jose and the tree as shown in the figure. Lets use cosine to find h in Paul's trianlge:
$cos(59)= \frac{19}{h}$
$h= \frac{19}{cos(59)}$
$h=36.9$
Now we can use the law of sines to find the distance $x$ between Paul and Jose:
$\frac{sin(43)}{36.9} = \frac{sin(16)}{x}$
$x= \frac{36.9sin(16)}{sin(43)}$
$x=14.9$

Now that we know the distance between Paul and Jose, the only thing left is add that distance to the distance from Paul and the base of the tree:
$19m+14.9=33.9m$

We can conclude that Jose is 33.9m from the base of the tree.

3 0

3 years ago

An explosion causes debris to rise vertically with an initial speed of 120 feet per second. The formula h equals negative 16 t s

Novay_Z [31]

Answer:

The debris will be at a height of 56 ft when time is <u>0.5 s and 7 s.</u>

Step-by-step explanation:

Given:

Initial speed of debris is, $s=120\ ft/s$

The height 'h' of the debris above the ground is given as:

$h(t)=-16t^2+120t$

As per question, $h(t)=56\ ft$ . Therefore,

$56=-16t^2+120t$

Rewriting the above equation into a standard quadratic equation and solving for 't', we get:

$-16t^2+120t-56=0\\\textrm{Dividing by -8 throughout, we get}\\\frac{-16}{-8}t^2+\frac{120}{-8}t-\frac{56}{-8}=0\\2t^2-15t+7=0$

Using quadratic formula to solve for 't', we get:

$t=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\t=\frac{-(-15)\pm \sqrt{(-15)^2-4(2)(7)}}{2(2)}\\\\t=\frac{15\pm \sqrt{225-56}}{4}\\\\t=\frac{15\pm\sqrt{169}}{4}\\\\t=\frac{15\pm 13}{4}\\\\t=\frac{15-13}{4}\ or\ t=\frac{15+13}{4}\\\\t=\frac{2}{4}\ or\ t=\frac{28}{4}\\\\t=0.5\ s\ or\ t=7\ s$

Therefore, the debris will reach a height of 56 ft twice.

When time $t=0.5\ s$ during the upward journey, the debris is at height of 56 ft.

Again after reaching maximum height, the debris falls back and at $t=7\ s$ , the height is 56 ft.

5 0

3 years ago

Which statement is true of triangles FGH and JKL?

Mars2501 [29]

Answer:

They are similar because their corresponding angles are congruent and their corresponding sides are proportional in length.

Step-by-step explanation:

From the statements,

∠G≅∠K

∠F≅∠J

∠H≅∠L

The sides of triangle FGH are proportional to those of triangle JKL

FG/JK =GH/KL=FH/JL =1/2

6 0

3 years ago

Read 2 more answers

Pls help with this!!

ANEK [815]

A)
Let X be the food expenditure of a family.
Let the population mean of an expenditure of a family be.
Let the population standard deviation of an expenditure of a family be.
The proportion families spend more than $30 but less than $490 per month on food is,

7 0

3 years ago